Electromagnetic radiation prevalence

  • Thread starter Waldheri
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  • #1
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Hello,

Perhaps my wording is a bit off so excuse me for that. I think it best to explain by way of a thought experiment. Say there is a device that counts the amount of photons with the same wavelength for all possible wavelengths. What is then the intensity vs. wavelength curve that I would see if I put this device in space, and will it be significantly different from what I would see if I put it on earth?

I've tried searching for this but I think I use the wrong terms in Google to get what I want. I will be very thankful for anyone that can provide some good sources on this.

Cheers,
Waldheri
 

Answers and Replies

  • #2
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If you're pointing the detector at some random point in space, then the spectrum you will see just depends on what celestial bodies are in the region of space that the detector can see, and what type of radiation these bodies emit. If you're pointing it at the sun then you will see the spectrum of the sun. http://en.wikipedia.org/wiki/Sunlight

On earth, you need to consider the earth's atmosphere both absorbing and scattering the light. That page and also the page on Rayleigh scattering are relevant to the processes for sunlight. In terms of other types of radiation from space, I'm not sure which are scattered or absorbed, but if you search for the answer to that question, it should be easier to find.
 
  • #3
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Thank you for your reply.

I asked this questions because I was thinking of the human eye, and how it can only detect a small slice in the electromagnetic spectrum. First I thought this had to mean we are blind to most em-radiation, but upon second thought I realized I also need to know how much em-radiation with the visible wavelengths there actually isin comparison to other em-radiation.

So it boils down to: we see only a tiny slice of the spectrum, but most radiation is in that spectrum or do we actually miss the majority of what's to be seen?
 
  • #4
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It's definitely the case that our eyes simply can't detect other wavelengths, regardless of how much radiation is present in those parts of the spectrum. But in terms of the amounts, the page for infrared says "Sunlight at zenith provides an irradiance of just over 1 kilowatt per square meter at sea level. Of this energy, 527 watts is infrared radiation, 445 watts is visible light, and 32 watts is ultraviolet radiation." So even neglecting all the rest of the spectrum, we are missing over half of what reaches us from the sun.

I don't really know the answer to your general question though, obviously.
 

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